hi everyone, I'm new so I hope I am doing this right.

I don't know how to solve this problem:

Use the algebraic method to write √2 sin x - √2 cos x in the form kcos(x + ϕ) where k>0, 0≤ ϕ ≤ 2π

thanks for any help!

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- Sep 29th 2007, 08:55 AMNeonhelp with trig equations
hi everyone, I'm new so I hope I am doing this right.

I don't know how to solve this problem:

Use the algebraic method to write √2 sin x - √2 cos x in the form kcos(x + ϕ) where k>0, 0≤ ϕ ≤ 2π

thanks for any help! - Sep 29th 2007, 09:39 AMred_dog
$\displaystyle \sqrt{2}\sin x-\sqrt{2}\cos x=-2\left(\frac{\sqrt{2}}{2}\cos x-\frac{\sqrt{2}}{2}\sin x\right)=$

$\displaystyle =-2\left(\cos x\cos\frac{\pi}{4}-\sin x\sin\frac{\pi}{4}\right)=-2\cos\left(x+\frac{\pi}{4}\right)$